A 5.00-kg howler monkey is swinging due east on a vine. It overtakes and grabs onto a 6.00-kg monkey also moving east on a second vine. The first monkey is moving at 10.0 m/s at the instant it grabs the second, which is moving at 7.00 m/s. 1) After they join on the same vine, what is their common speed
Question
Answer:
8.36 m/s
Explanation:
From the law of conservation of momentum,
Total momentum of the monkeys before grab = Total momentum after grab
mu+m’u’ = V(m+m’)………………. Equation 1
Where m = mass of the first money, m’ = mass of the second monkey, u = initial velocity of the first monkey, u’ = initial velocity of the second monkey, V = Common velocity of both monkeys.
make V the subject of the equation
V = [mu+m’u’]/(m+m’)……………. Equation 2
Given: m = 5 kg, m’ = 6 kg, u = 10 m/s, u’ = 7 m/s
Substitute into equation 2
V = [5×10+6×7]/(5+6)
V = [50+42]/11
V = 92/11
V = 8.36 m/s